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If $V$ is the natural representation for $\operatorname{GL}(2,q)$, then $V⊗V$ appears to decompose into the direct sum of a (strange?) one-dimensional module and a three dimensional module. I've managed to confuse myself about where $\operatorname{Sym}(V)$ and $Λ(V)$ come into play, and maybe some duals are messed up. I am also interested in taking those 1 and 3 dimensional modules and dividing them by the determinant.

Can someone explain the 1, 2, and 3 dimensional modules of $\operatorname{GL}(2,q)$ using the natural module, symmetric powers, exterior powers, and duals?

Some specific questions that if the answer is no, might require a nice gentle answer as to sane formulas for both sides of the inequality:

  • Does $V^*⊗V^*$ decompose into $\operatorname{Sym}(V^*)⊕1$, where is the trivial module?
  • Is $W/\det = W^*$, where $W/\det$ is the representation where $g$ acts as $g$ divided by its determinant (taken in $\operatorname{GL}_2$).
  • Does it make sense to talk about the determinant after one thinks of $g$ as acting on $W$? That is, as an $n×n$ matrix, rather than a $2×2$?
  • Is the action of $\operatorname{GL}(2,q)$ on polynomials (by change of variable) called $\operatorname{Sym}(V)$ or $\operatorname{Sym}(V^*)$?
  • What about the action on wedgies, $Λ(V)$ or $Λ(V^*)$?

If it is easier, I think I only care about $\operatorname{PGL}$ for today, but I wouldn't be surprised if I liked $\operatorname{GL}$ tomorrow. This confusion is all probably due to sticking to $\operatorname{SL}(2,q)$ where $W =W^*$ and the determinant is 1 so it doesn't really matter when you divide by it.

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  • $\begingroup$ What is $W$? An arbitrary representation? $\endgroup$ – Jim Belk Jun 3 '11 at 2:16
  • $\begingroup$ @Jim: Yes. Finite dimensional. And by finite, you can stick to dimension 1,2,3. I think I used n=dim(W). $\endgroup$ – Jack Schmidt Jun 3 '11 at 2:24
  • $\begingroup$ Another (phrasing of the same) question: The submodule of V⊗V generated by { v⊗v : v in V } is three dimensional right? Basis given by { e1⊗e1, e2⊗e2, e1⊗e2 + e2⊗e1 }. Are these the polynomials { xx, yy, xy/2 } or do I need to fix a determinant? $\endgroup$ – Jack Schmidt Jun 3 '11 at 2:30
  • $\begingroup$ Dear Jack, Because you asked about $G(2,q)$, rather than say $GL_2(\mathbb C)$, I presumed your question was about (possible) divergences between char. 0 and char. p. I evidently misunderstood, and so have deleted my answer. Regards, $\endgroup$ – Matt E Jun 3 '11 at 2:54
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I have some answers but not others. Also, I am frightened of finite fields, so I will only talk about representations over $\mathbb{C}$.

  1. The tensor product $V\otimes V\;$ is the direct sum of the alternating square $\Lambda^2(V)$ (which is one-dimensional) and the symmetric square $\text{Sym}_2(V)$ (which is three-dimensional).

  2. The action of $\mathrm{GL}(2,\mathbb{C})$ on $\Lambda^2(V)$ is multiplication by determinants, i.e. $gx = \det(g)x$ for any real number $x$. The action on the dual $\Lambda^2(V^*)$ is division by determinants.

    Note: By $\Lambda^2(V)$ here I mean the second exterior power of $V$, i.e. linear combinations of wedge products $v \land w$ where $v,w\in V$. The action is $g(v\land w) = (gv)\land (gw)$, e.g. $$ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}(e_1\land e_2) \;=\; (2e_1) \land (3e_2) \;=\; 6(e_1\land e_2). $$

  3. The action of $\mathrm{GL}(2,\mathbb{C})$ on $\text{Sym}_2(V)$ can be thought of as the action of $\mathrm{GL}(2,\mathbb{C})$ on the space of quadratic forms on $\mathbb{C}^2$ via $q \mapsto g q g^T$. This generalizes to polynomials.

  4. The action of $\mathrm{GL}(2,\mathbb{C})$ on $\text{Sym}_2(V^*)$ can be thought of as the action of $\mathrm{GL}(2,\mathbb{C})$ on the space of quadratic forms on $\mathbb{C}^2$ via $q \mapsto (g^{-1})^T q g^{-1}$. This generalizes to polynomials.

Note: After thinking about it, I decided to remove the names "change of basis" and "change of variables" on #3 and #4, since I'm not sure which one you mean when you say "change of variables".

Difference Between #3 and #4: This is a bit complicated, because there are actually four different ways to change variables with a matrix. For example, given the matrix $\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$ with variables $x$ and $y$, we can define new variables $u$ and $v$ in any one of the following ways:

A. $\begin{bmatrix}u \\ v\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$

B. $\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix}$

C. $\begin{bmatrix}u & v\end{bmatrix} = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$

D. $\begin{bmatrix}x & y\end{bmatrix} = \begin{bmatrix}u & v\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$

This leads to four different actions of matrices on polynomials. Of these four actions, two are left actions (namely those derived from methods A and D), and two are right actions (namely those derived from methods B and C). For left actions, applying $g$ first and then $h$ is the same as applying $hg$: $$ h(gp) \;=\; (hg)p. $$ (For right actions, applying $g$ and then applying $h$ would be the same as applying $gh$). Since we are only interested in left actions, this leaves us with two ways that a matrix might act on a polynomial. One of these (method D) gives the action of matrices on $\text{Sym}_2(V)$ while the other (method A) gives the action of matrices on $\text{Sym}_2(V^*)$.

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  • $\begingroup$ Can you explain the difference between #3 and #4? If it can be said using polynomials, I'll understand that better today, though I'll think about the quadratic forms version (with your very explicit action) over the next day. #2 contradicts my calculations, so I'm still confused about something. $\endgroup$ – Jack Schmidt Jun 3 '11 at 2:41
  • $\begingroup$ I'll add some clarifying edits. $\endgroup$ – Jim Belk Jun 3 '11 at 3:00
  • $\begingroup$ #2 has helped! Since Λ(V) is one-dimensional, its dual (as in, inverse transpose) is definitely dividing by the determinant, and so W⊗Λ(V)* = W/det. This might explain some of the dual vs. determinant confusion. V*⊗V* = Sym(V)*⊕Λ(V)* seems right, but I was also thinking V*⊗V* = (V⊗Λ(V)*)⊗(V⊗Λ(V)*), which I now think is nonsense. Instead (V⊗Λ(V)*)⊗(V⊗Λ(V)*) = V⊗V⊗(Λ(V)*)^2, is now dividing by the determinant squared. When I tried this over a field of size 3 for instance, it cancelled, and then over size 5 it did not. $\endgroup$ – Jack Schmidt Jun 3 '11 at 3:47
  • $\begingroup$ My main confusion left is which one is "polynomials", Sym(V) or Sym(V*), and I think this must come down to being able to see the difference between the verbal description of #3 and #4 (the explicit action is clear, but depending on which side things act on I could make the transposes and the inverses move around in all sorts of confusing ways, so I prefer the verbal). $\endgroup$ – Jack Schmidt Jun 3 '11 at 3:48
  • $\begingroup$ Marvelous. To me these are all sort of covered by "change of variables", and seeing them listed like this makes it very easy to very clearly distinguish them. In the PGL(2,q) case I still have some extra isomorphisms I don't understand (V⊗V=V*⊗V*), but I think GL(2,C) is making some sense. $\endgroup$ – Jack Schmidt Jun 3 '11 at 5:14

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